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Let $\mathbb{B}_r$ denote the closed disk $\{z \in \mathbb{C} : |z| \le r \}$. State whether $\infty$ is a removable singularity , pole , or essential singularity in the following statement

$f$ is an entire function for which $f^{-1}(B_r)$ is bounded for all $r >0$

My attempt :Here $\infty$ are both removable singularity and pole

Inverse image of closed unit disk of $f$ is bounded so $f$ can not have essential singularity at $\infty$

Removable singularity :-

Suppose that $f$ is an entire function that has a removable singularity at infinity.Then there exist an entire function $g$ such that $g(z)= f(1/z)$ for all $z \in \mathbb{B_r}-\{0\}$.This implies that $\lim_{z\to \infty} f(1/z)=f(0) $ which in turn implies that $g$ is bounded.Since g is a bounded entire function, by liouville's theorem , $g$ is constant.Hence $ f$ is constant $\implies f^{-1}(B_r)$ is bounded for all $r >0$

For pole

take $f(z)= z$ , $g(z)=f(1/z)=1/z$ where $|z|\le r >0$. $g(z)$ has pole at $z=0 \implies f$ has a pole at $\infty$ $\implies f^{-1}(B_r)$ is bounded for all $r >0$

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Your second example is fine. For removable singularity the argument is simpler. Any entire function with removable singularty at $\infty$ is a constant by Liouville's Theorem because it is bounded.

If $f$ has an essential singularity at $\infty$ then it would take values in $B_1$ in every neighborhood of $\infty$ so $f^{-1} (B_1)$ is not bounded. Hence, $f$ cannot have an essential singularity at $\infty$.

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    $\begingroup$ @wasiu If it is a removable singularity then $f$ is a consatnt. So $f^{-1}(B_r)=\mathbb C$ for large enough $r$ , contradicting the hypothesis. $\endgroup$ May 26, 2022 at 4:45
  • $\begingroup$ okay got it ,thanks you @Kavi sir $\endgroup$
    – wasiu
    May 26, 2022 at 4:48

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